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mngnmfn · Nov 6, 2010, 12:37 PM

I'm beyond lost here, is there a way to prove that?

galactus · Nov 6, 2010, 02:53 PM

$sec^{2}(x)=\frac{csc(x)}{csc(x)-sin(x)}$

Start with the right hand side:

Since $csc(x)=\frac{1}{sin(x)}$ :

$\frac{\frac{1}{sin(x)}}{\frac{1}{sin(x)}-sin(x)}$

$\frac{\frac{1}{sin(x)}}{\frac{1-sin^{2}(x)}{sin(x)}}$

$\frac{1}{sin(x)}\cdot \frac{sin(x)}{1-sin^{2}(x)}$

sin(x) cancels and we are left with:

$\frac{1}{1-sin^{2}(x)}$

Remember that $cos^{2}(x)=1-sin^{2}(x)$

$\frac{1}{cos^{2}(x)}=sec^{2}(x)$

mngnmfn · Nov 6, 2010, 03:02 PM

Thank you! You're brilliant!